Please help what is the answer?

Unit 8 right triangles and trigonometry homework 3 similar right triangles and geometric mean

Ugo [173]

The right triangles that have an altitude which forms two right triangles

are similar to the two right triangles formed.

Responses:

1. ΔLJK ~ ΔKJM

ΔLJK ~ ΔLKM

ΔKJM ~ ΔLKM

2. ΔYWZ ~ ΔZWX

ΔYWZ ~ ΔYZW

ΔZWX ~ ΔYZW

3. x = 4.8

4. x ≈ 14.48

5. x ≈ 11.37

6. G.M. = 12·√3

7. G.M. = 6·√5

<h3>What condition guarantees the similarity of the right triangles?</h3>

1. ∠LMK = 90° given

∠JMK + ∠LMK = 180° linear pair angles

∠JMK = 180° - 90° = 90°

∠JKL ≅ ∠JMK All 90° angles are congruent

∠LJK ≅ ∠LJK reflexive property

ΔLJK is similar to ΔKJM by Angle–Angle, AA, similarity postulate

∠JLK ≅ ∠JLK by reflexive property

ΔLJK is similar to ΔLKM by AA similarity

By the property of equality for triangles that have equal interior angles, we have;

ΔKJM ~ ΔLKM

2. ∠YWZ ≅ ∠YWZ by reflexive property

∠WXZ ≅ ∠YZW all 90° angle are congruent

ΔYWZ is similar to ΔZWX, by AA similarity postulate

∠XYZ ≅ ∠WYZ by reflexive property

∠YXZ ≅ ∠YZW all 90° are congruent

ΔYWZ is similar to ΔYZW by AA similarity postulate

Therefore;

ΔZWX ~ ΔYZW

3. The ratio of corresponding sides in similar triangles are equal

From the similar triangles, we have;

$\dfrac{8}{10} = \mathbf{ \dfrac{x}{6}}$

8 × 6 = 10 × x

48 = 10·x

$x = \dfrac{48}{10} = \underline{4.8}$

3. From the similar triangles, we have;

$\mathbf{\dfrac{20}{29}} = \dfrac{x}{21}$

20 × 21 = x × 29

420 = 29·x

$x = \dfrac{420}{29 } \approx \underline{14.48}$

4. From the similar triangles, we have;

$\mathbf{\dfrac{20}{52}} = \dfrac{x}{48}$

20 × 48 = 52 × x

$x = \dfrac{20 \times 48}{52} = \dfrac{240}{13} \approx \underline{18.46}$

5. From the similar triangles, we have;

$\mathbf{\dfrac{13.2}{26}} = \dfrac{x}{22.4}$

13.2 × 22.4 = 26 × x

$x = \dfrac{13.2 \times 22.4}{26} \approx \underline{ 11.37}$

6. The geometric mean, G.M. is given by the formula;

$G.M. = \mathbf{\sqrt[n]{x_1 \times x_2 \times x_3 ... x_n}}$

The geometric mean of 16 and 27 is therefore;

$G.M. = \sqrt[2]{16 \times 27} = \sqrt[2]{432} = \sqrt[2]{144 \times 3} = \mathbf{12 \cdot \sqrt{3}}$

The geometric mean of 16 and 27 is 12·√3

7. The geometric mean of 5 and 36 is found as follows;

$G.M. = \sqrt[2]{5 \times 36} = \sqrt[2]{180} = \sqrt[2]{36 \times 5} = \mathbf{ 6 \cdot \sqrt{5}}$

The geometric mean of 5 and 36 is 6·√5

Learn more about the AA similarity postulate and geometric mean here:

brainly.com/question/12002948

brainly.com/question/12457640

7 0

2 years ago

If 15 times a number to equal 647

Ainat [17]

I'm assuming, from what you've given me that the question is asking to do this.

15n=647
647 divided by 15 equals 43.13...repeated
Let me know if this is what you are looking for.

3 0

3 years ago

If the third of 6 consecutive numbers is 12, what is their sum?

Oksana_A [137]

Answer:

75

Step-by-step explanation:

If the third of 6 consecutive numbers is 12, what is their sum?

6 numbers = c-2, c-1, c, c+1, c+2, and c+3

The numbers become 10, 11, 12, 13, 14, and 15

Add that together and you get 75.

6 0

3 years ago

A university wants to build a new residence hall with dorm rooms. The length is 3x+1 ft and the width is x^2-1 ft. If the univer

Masja [62]

Length of room, l = 3x + 1 ft.

Breadth of room, b = x² -1 ft.

Now, it is given that university wants each room to have 195 ft² of living space.

So,

$lb = A\\\\(x^2-1)(3x+1)=195\\\\3x^3+x^2-6x-1=195\\\\3x^3+x^2-6x-196=0$

So, above equation has two complex root and one real root i.e x = 4.08 ft .

Therefore, Length of room is 13.24 ft and breadth is 15.65 ft.

Hence, this is the required solution.

7 0

3 years ago

Find sin, cos, and tan for the given point (-4,5).

grandymaker [24]

Step-by-step explanation:

x^2 + y^2 = r^2 --> r = $\sqrt{(-4)^2 + 5^2}$ = $\sqrt{41}$

cos = x /r = -4 / $\sqrt{41}$

sin = y /r = 5/ $\sqrt{41}$

tan = sin /cos = -5 /4

7 0

3 years ago

Please help what is the answer?​

Please help what is the answer?